Suppose we take a random sample of size n from a normal population with variance, σ2 . It can be shown that (n−1)s2/σ2 has a chi-square distribution with n−1 degrees of freedom, where s is the sample variance. Below is a random sample of size 8 drawn from a normal population. Use the above fact to compute a 95% confidence interval for the population variance, σ2 .
Sample 1: 11.6, 17.2, 15.0, 16.3, 22.9, 13.5, 16.4, 16.1
Below is a second random sample, independent from the first, of size 8 from a second normal population. Remembering that the F distribution is a ratio of independent chi- squares divided by their degrees of freedom, it can be shown that, under random, independent sampling, if the variances of the populations are equal, then s21/s2 has an F distribution with, in this case, 7 numerator and 7 denominator degrees of freedom (where the degrees of freedom are n − 1 for the corresponding samples). Test at α = .05 the null hypothesis that the variances are equal against the alternative that the variance of the first population is greater.
Sample 2: 17.7, 11.0, 17.0, 12.4, 10.8, 9.9, 17.2, 10.1
For your first problem:
Standard Deviation = 3.29 (Standard deviation is the square root of the variance)
Variance = 10.82 (Variance is standard deviation squared)
Using a chi-square table for the endpoints:
(n-1)s^2/16 to (n-1)s^2/1.69
(8-1)10.82/16 to (8-1)10.82/1.69
7(10.82)/16 to 7(10.82)/1.69
75.74/16 to 75.74/1.69
4.73 to 44.82 -->confidence interval for the variance
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For the second problem:
Standard Deviation = 3.43
Variance = 11.76
Sample 1: n = 8; variance = 10.82; df = n - 1 = 7
Sample 2: n = 8; variance = 11.76; df = n - 1 = 7
Test statistic = sample 1 variance / sample 2 variance
You can use the F-distribution at .05 level using the above information for degrees of freedom. This will be your critical value to compare to the test statistic. If the test statistic exceeds the critical value from the table, the null will be rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null is not rejected.
I'll let you take it from here to finish. Check these calculations!
To compute a 95% confidence interval for the population variance, σ2, based on the given sample (Sample 1), we need to use the chi-square distribution.
1. Calculate the sample variance (s2) for Sample 1:
- Compute the mean of the sample values: (11.6 + 17.2 + 15.0 + 16.3 + 22.9 + 13.5 + 16.4 + 16.1) / 8 = 15.575
- Calculate the sample variance: s2 = Σ(xi - x̄)2 / (n - 1)
(11.6 - 15.575)2 + (17.2 - 15.575)2 + ... + (16.1 - 15.575)2 / (8 - 1)
= 7.5875
2. Determine the critical values from the chi-square distribution:
- Since we want a 95% confidence interval, the alpha level (α) is equal to 1 - 0.95 = 0.05.
- The degrees of freedom (df) are calculated as n - 1 = 8 - 1 = 7.
- The critical values for a chi-square distribution with 7 degrees of freedom and an alpha level of 0.05 can be found using statistical tables or software. For this example, the critical values are approximately 2.167 and 17.535.
3. Compute the confidence interval:
- The confidence interval for the population variance, σ2, is given by the formula:
[(n-1)s2 / critical value, upper , (n-1)s2 / critical value, lower ]
[(7)(7.5875) / 17.535, (7)(7.5875) / 2.167]
[0.894, 24.769]
Therefore, the 95% confidence interval for the population variance, σ2, based on Sample 1, is [0.894, 24.769].
Now, let's move on to the second part of the question.
To test the null hypothesis that the variances are equal against the alternative that the variance of the first population is greater (Sample 1 vs. Sample 2), we need to use the F distribution.
4. Calculate the sample variances (s21 and s22) for both Sample 1 and Sample 2:
- For Sample 1, we already calculated s2 = 7.5875.
- For Sample 2:
- Compute the mean of the sample values: (17.7 + 11.0 + 17.0 + 12.4 + 10.8 + 9.9 + 17.2 + 10.1) / 8 = 13.525
- Calculate the sample variance: s2 = Σ(xi - x̄)2 / (n - 1)
(17.7 - 13.525)2 + (11.0 - 13.525)2 + ... + (10.1 - 13.525)2 / (8 - 1)
= 10.6700
5. Determine the critical value from the F distribution:
- Since the degrees of freedom for both numerator (Sample 1) and denominator (Sample 2) are 7, the critical value can be found using statistical tables or software. For an alpha level of 0.05, the critical value is approximately 3.682.
6. Compute the test statistic:
- Calculate the F test statistic: F = s21 / s22
= 7.5875 / 10.6700
= 0.7111
7. Compare the test statistic with the critical value:
- If the test statistic is greater than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.
Since 0.7111 < 3.682, we fail to reject the null hypothesis.
Therefore, at α = 0.05, there is not enough evidence to conclude that the variance of the first population is greater than the variance of the second population.